OCR ADVANCING PHYSICS B G494 EXAM MONDAY 18th June

Why would object appears weightless if earth spins faster !??

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Reply 62

Original post by xiyangliu

Why would object appears weightless if earth spins faster !??

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i think because if the force due to circular motion (centripetal force) is high enough, (by either making the radius smaller or increasing the velocity of the spin: F=mv^2/r) it will equal the force on a mass due to the gravitational field strength g. so the two forces will cancel out giving a resultant of 0.
it might be centrifugal, but yeah... something along these lines i think.

Reply 63

Original post by Dhanu123

i think because if the force due to circular motion (centripetal force) is high enough, (by either making the radius smaller or increasing the velocity of the spin: F=mv^2/r) it will equal the force on a mass due to the gravitational field strength g. so the two forces will cancel out giving a resultant of 0.
it might be centrifugal, but yeah... something along these lines i think.

But I thought the centrifugal force ? Never taught that before. Where does it come from ?

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Reply 64

Original post by xiyangliu

But I thought the centrifugal force ? Never taught that before. Where does it come from ?

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centrifugal force is not actually a force, but it acts as a reaction to the centripetal force. im not really sure about this topic that much either. i need to revise it a bit more

Reply 65

Original post by Dhanu123

centrifugal force is not actually a force, but it acts as a reaction to the centripetal force. im not really sure about this topic that much either. i need to revise it a bit more

Is it in the syllabus ?? What chapter ??

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Reply 66

freakymonkey

4

Original post by Dhanu123

how to do you calculate rate of collision of particles against a flat surface area (or wall)?

In these styles of question you'll usually be made to work out the pressure exerted on the wall (p=F/A, so pressure = (Number of Molecules * Force exerted on wall per molecule, which is 2mv)/Area of wall) first. I've never seen rate of collision for the entire wall, but I have seen rate of collision of one particle across a sphere - 1 mark for realising it travelled 4*the radius between hitting each point, and then realising that number of times = v/4r.

Does this answer any questions. My explaining skills are... limited, so if this raises anything let me know :smile:

Reply 67

Can someone help me work out question 7 (a) of Jan 2012?

"7 The rate of rotation of a distant spiral galaxy, like that shown in Fig. 7.1, can be found by comparing
the light from the left and right hand side of the galaxy.

(a) Explain why there will be a difference in the redshift of the light from the left and right hand
sides of the galaxy"

I don't understand why there's a difference.

Reply 68

luketaylor777

Original post by Rotravis

Can someone help me work out question 7 (a) of Jan 2012?

"7 The rate of rotation of a distant spiral galaxy, like that shown in Fig. 7.1, can be found by comparing
the light from the left and right hand side of the galaxy.

(a) Explain why there will be a difference in the redshift of the light from the left and right hand
sides of the galaxy"

I don't understand why there's a difference.

Overall the galaxy will show redshift as it is moving away, but due to its rotation, one side of the galaxy with be rotating towards us so it's light will show less redshift than the light at the other side of the galaxy. That's my guess anyway

(edited 11 years ago)

Reply 69

Original post by luketaylor777

Overall the galaxy will show redshift as it is moving away, but due to its rotation, one side of the galaxy with be rotating towards us so it's light will show less redshift than the light at the other side of the galaxy. That's my guess anyway

Ah cool that fits the mark scheme!

Question 10 of jan 2012 (c) doesn't work for me and I'm sure i'm doing it right.

It's supposed to be really easy but I'm not getting the correct answer. Anyone know why? I'm just subbing values into the equation in B

Reply 70

Original post by Rotravis

Ah cool that fits the mark scheme!

Question 10 of jan 2012 (c) doesn't work for me and I'm sure i'm doing it right.

It's supposed to be really easy but I'm not getting the correct answer. Anyone know why? I'm just subbing values into the equation in B

remember to square T (8.6 * 10^4 all squared) and thats its only pi that's squared. also cube root at the end. there shouldn't be an error, unless you typed the numbers in wrong. you should get 4.2 * 10^7 :smile:

Reply 71

Original post by Dhanu123

remember to square T (8.6 * 10^4 all squared) and thats its only pi that's squared. also cube root at the end. there shouldn't be an error, unless you typed the numbers in wrong. you should get 4.2 * 10^7 :smile:

I've tried it about 10 times now. r^3 = (6.7x10^-11) x (6x10^24) x (8.6x10^4)^2 all over 4pi^2.

I get 7.33x10^24 before the cube root.. after cube root I get 19.4x10^7m :frown:

Reply 72

Original post by Rotravis

I've tried it about 10 times now. r^3 = (6.7x10^-11) x (6x10^24) x (8.6x10^4)^2 all over 4pi^2.

I get 7.33x10^24 before the cube root.. after cube root I get 19.4x10^7m :frown:

i get 7.53*10^22 before the cube root, see if you miss typed anything, or if you missed a bracket somewhere :smile:

Reply 73

Original post by Dhanu123

i get 7.53*10^22 before the cube root, see if you miss typed anything, or if you missed a bracket somewhere :smile:

Ugh I tried it with 4pi^2 first then did the top/ANS and it worked. Bit weird that I could only get it to work like that but meh.

Reply 74

Original post by Dhanu123

how to do you calculate rate of collision of particles against a flat surface area (or wall)?

you usually get given something like the distance between the collisions. Then you divide the total d/d between collisions to estimate no of collisions.

Reply 75

Original post by Rotravis

Ugh I tried it with 4pi^2 first then did the top/ANS and it worked. Bit weird that I could only get it to work like that but meh.

I got the following sum:

r^3=GM(T^2)/4pi^2 from previous q

r^3= (6.7*10^-11 (G) x 6*10^24 (M) x ((8.6*10^4)^2) (T^2))/(4*pi^2)

(be careful with your calculator, use brackets where appropriate.

r^3= 7.53*10^22

r= cube root of (7.53*10^22)

r= 4.2*10^7

I am really unsure if I will be calm enough to do this kind of thing in the actual exam under time pressure.

(edited 11 years ago)

Reply 76

The factor k^2 is also the ratio of the frequency of a trap radar signal reflected back from a car coming towards the speed trap, to the frequency of the signal transmitted towards the car. If the speed trap equipment can detect a change in frequency of 1 part in 10^8, what is the smallest car speed it can detect ?? Given v/c = k^2-1/k^2+1. Wtf ....

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Reply 77

Quantum_Phantom

7

Original post by PhysicsGirl

Okay, so now G495 is out of the way, roll on Rise and Fall of The Clockwork Universe. Any revision materials/ tips / questions/ discussions of past papers welcome :-)

Okay, so following the suggestion of Oromis (the physics revision God of TSR, clearly :-P [It's okay for me to take the mick, I know him]) I'm updating this post with the revision materials/ links that I've put up so far, and I'll continue to do so when other people add things in- just draw my attention (via quoting) to something you think should be posted up here and I'll edit it in.

So, here again is the link to the Matthew Arnold Moodle; http://83.170.93.38/~matthew/course/view.php?id=62 for all those past paper materials and mark schemes, as well as some other resources (Choose guest access if it asks you to log in)

Attached to the post is my revision checklist for the entire exam (adapted from the specification, so hopefully is quite complete). Also attached are my notes for Chapter 10.

And here's the list of chapters/ topics for the exam:
Chapter 10: Creating Models (oscillators, SHM, capacitance, advantages and disadvantages of modelling techniques)
Chapter 11: Out Into Space (gravity, circular motion (inc. orbits) forces and momentum)
Chapter 12: Our Place in the Universe (relativity, Doppler shift, Big Bang Theory, gamma factor, distances in space)
Chapter 13: Matter: Very Simple (gas laws, kinetic model of particles)
Chapter 14: Matter: Very Hot and Very Cold (energy, activation processes, Boltzmann)

Have you got notes for the other chapters ?

Thanks in advance

Reply 78

Original post by xiyangliu

The factor k^2 is also the ratio of the frequency of a trap radar signal reflected back from a car coming towards the speed trap, to the frequency of the signal transmitted towards the car. If the speed trap equipment can detect a change in frequency of 1 part in 10^8, what is the smallest car speed it can detect ?? Given v/c = k^2-1/k^2+1. Wtf ....

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Where is this question from?

This question has an uncanny resemblance to another question I can sort of think of.

But consider the following ratio:

dλ/λ = v/c

Reply 79